Summation and series: RI 2024 Promo Q5

(a)[3] $$\begin{aligned} \sum_{r=0}^{n}[(n+2)r+n] &=\frac{n+1}{2}[n+n(n+2)+n]\\ &=\frac{n+1}{2}(n^2+4n)\\ &=\frac n2(n+1)(n+4). \end{aligned}$$ Alternative $$\begin{aligned} \sum_{r=0}^{n}[(n+2)r+n] &=\sum_{r=0}^{n}(n+2)r+\sum_{r=0}^{n}n\\ &=(n+2)\sum_{r=1}^{n}r+(n+1)n\\ &=(n+2)\cdot\frac n2(n+1)+(n+1)n\\ &=\frac n2(n+1)(n+4). \end{aligned}$$ (b)[3] $$\begin{aligned} \sum_{r=1}^{n}(r+2)^3&=3^3+4^3+\cdots+(n+1)^3+(n+2)^3\\ &=\sum_{r=3}^{n+2}r^3\\ &=\sum_{r=1}^{n+2}r^3-1^3-2^3\\ &=\frac14(n+2)^2(n+3)^2-9. \end{aligned}$$ (c)[3] $$\begin{aligned} 1^3-2^3+3^3-4^3+5^3-6^3+\cdots+(2n-1)^3-(2n)^3 &=\sum_{r=1}^{2n}r^3-2\sum_{r=1}^{n}(2r)^3\\ &=\sum_{r=1}^{2n}r^3-16\sum_{r=1}^{n}r^3\\ &=\frac14(2n)^2(2n+1)^2-16\left[\frac14n^2(n+1)^2\right]. \end{aligned}$$

$$\begin{aligned} &=n^2(2n+1)^2-4n^2(n+1)^2\\ &=n^2[(2n+1)^2-(2n+2)^2]\\ &=-n^2(4n+3). \end{aligned}$$